Answer:
Option B
1025 psi
Explanation:
In a single shear, the shear area is
The shear strength= and in this case
Shear strength= hence making load the subject then
Load=Shear area X Shear strength
Load=
To test the effects of a new fertilizer, 100 plots were divided in half. Fertilizer A is randomly applied to one half, and B to
2 answers:
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a) Potential in A: -2700 V
b) Potential difference: -26,800 V
c) Work:
Explanation:
a)
The electric potential at a distance r from a single-point charge is given by:
where
is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.
Charge 1 is
and is located at the origin (x=0, y=0)
Charge 2 is
and is located at (x=0, y = 0.40 m)
Point A is located at (x = 0.40 m, y = 0)
The distance of point A from charge 1 is
So the potential due to charge 2 is
The distance of point A from charge 2 is
So the potential due to charge 1 is
Therefore, the net potential at point A is
b)
Here we have to calculate the net potential at point B, located at
(x = 0.40 m, y = 0.30 m)
The distance of charge 1 from point B is
So the potential due to charge 1 at point B is
The distance of charge 2 from point B is
So the potential due to charge 2 at point B is
Therefore, the net potential at point B is
So the potential difference is
c)
The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by
where
q is the charge of the particle
is the potential difference
In this problem, we have:
is the charge of the electron
is the potential difference
Therefore, the work required on the electron is